Question

A dielectric of dielectric constant 2.5 is filled in the gap between the plates of a capacitor. Calculate the factor by which the capacitance is increased, if the dielectric is only sufficient to fill up one-fourth of the gap.​

Answer 1

Explanation:

I hope it will help you...

Answer 2

⭐Assumption:-⭐

In case C, before filling The gap is d

after filling , In case C₁ gap is 3/4d andC₂ gap is 1/4d

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⭐Solution:-⭐

• Capacitor capacity until full

$\boxed{C= \dfrac{E_o.s}{d}}$

• Capacitor capacity after filling

$\boxed{C'=\dfrac{C_1.C_2}{C_1+C_2}}$-------------(2)

Where,

$C_1= \dfrac{E.E_o.S}{1/4.d} = \dfrac{E. 4. E_o.S}{d} = 4.E.C$

$C_2= \dfrac{E_o.S}{3/4.d} = \dfrac{ 4. E_o.S}{3d} = \dfrac{4}{3}.C$

Substitude the values in equation (2), we get

$C' = \dfrac {16.E}{12.E+4}. C = K.C${ let the value is K}

=> BY SOLVING K, we get

1.176 value

that is Capacitor capacity will increase 1.176 times